5.1.2含30°角的直角三角形的性质 课件 -2026-2027学年湘教版数学八年级上册

#湘教版数学八年级上册5.1.2含30°角的直角三角形的性质专项练习##知识点回顾**性质**：在直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)如果一个锐角等于$\boldsymbol{30^\circ}$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)那么它所对的直角边等于斜边的一半。几何语言：在$Rt\triangleABC$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则$\boldsymbol{\(BC=\)\dfrac\(\boldsymbol{12}\)AB}$。**逆用（判定）**：在直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)如果一条直角边等于斜边的一半\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)那么这条直角边所对的锐角等于$30^\circ$。✅解题要点1.前提：**必须是直角三角形**\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)才有这个结论；普通三角形不成立。2.分清：$30^\circ$角**对的直角边**是短直角边\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)是斜边一半；不是邻边。3.常结合上一节：直角三角形斜边上中线、等腰、等边三角形一起考。满分：100分时间：40分钟##一、选择题（每题4分\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)共20分）1.在$Rt\triangleABC$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则$AC$是（）A.斜边$AB$的一半B.斜边$AB$的2倍C.直角边$BC$的一半D.直角边$BC$的2倍2.$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)斜边$\(\(AB=\)10\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则$BC$长为（）A.$10$B.$5$C.$20$D.$\(2.5\)$3.在直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)一条直角边等于斜边的一半\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则这条直角边所对的角为（）A.$60^\circ$B.$45^\circ$C.$30^\circ$D.$90^\circ$4.$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)短直角边$\(AC=4\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则斜边$\(AB=\)$（）A.$2$B.$4$C.$8$D.$16$5.下列说法正确的是（）A.有一个角$30^\circ$的三角形\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$30^\circ$对边是另一边一半B.在直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$30^\circ$角所对直角边等于斜边一半C.直角三角形斜边等于短直角边一半D.任意三角形\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$30^\circ$对边等于斜边一半##二、填空题（每题4分\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)共20分）1.在直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$30^\circ$锐角所对的________等于斜边的一半。2.$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)斜边$\(AB=\)14$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则$\(BC=\)$________。3.$Rt\triangle$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)若一条直角边等于斜边的一半\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则这条直角边所对的角是________。4.$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(AC=6\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则斜边$\(AB=\)$________。5.在$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则另一个锐角$\angleB=$________。##三、基础解答题（每题8分\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)共32分）1.已知：在$Rt\triangleABC$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)斜边$\(AB=\)\(\boldsymbol{12}\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)求$BC$的长。2.已知：$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(BC=\)3$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(AB=\)6$。求证：$\angleA=30^\circ$。3.判断对错\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)并说明理由（1）三角形有一个角是$30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)那么$30^\circ$对的边是斜边的一半。（2）在$Rt\triangle$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$60^\circ$角对的直角边是斜边一半。4.已知：$Rt\triangleABC$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(\(AB=\)10\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)求$AC$的长。##四、拓展提升题（每题14分\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)共28分）1.已知：在$Rt\triangleABC$中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleC=90^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleA=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$CD$是斜边$AB$上的中线\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)求证：$\triangleBCD$是等边三角形。2.已知：$\angleAOB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)点$P$在$OA$上\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$PC\perpOB$于$C$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(PC=5\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)求$OP$的长度。---#参考答案与解析##一、选择题1.A2.B3.C4.C5.B##二、填空题1.**直角边**2.$\boldsymbol{\(\boldsymbol{7}\)}$3.$\boldsymbol{30^\circ}$4.$\boldsymbol{\(\boldsymbol{12}\)}$5.$\boldsymbol{60^\circ}$##三、基础解答题1.解：$\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)\angleA=30^\circ$$\therefore\(BC=\)\dfrac\(\boldsymbol{12}\)AB$（直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$30^\circ$角所对直角边等于斜边一半）$\because\(AB=\)\(\boldsymbol{12}\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\therefore\(BC=\)\boldsymbol{6}$2.证明：$\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)\(BC=\)3\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)\(AB=\)6$$\therefore\(BC=\)\dfrac\(\boldsymbol{12}\)AB$$\therefore\angleA=30^\circ$（直角三角形中\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)如果一条直角边等于斜边一半\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)则这条直角边所对锐角是$30^\circ$）3.（1）×；理由：必须是**直角三角形**才有这个结论\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)普通三角形不成立。（2）×；理由：$30^\circ$角所对直角边等于斜边一半\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)不是$60^\circ$。4.解：$\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)\angleB=30^\circ$$\thereforeAC=\dfrac\(\boldsymbol{12}\)AB$$\because\(\(AB=\)10\)$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\thereforeAC=\boldsymbol{5}$##四、拓展提升题1.证明：$\because\angleACB=90^\circ\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)\angleA=30^\circ$$\therefore\(BC=\)\dfrac\(\boldsymbol{12}\)AB$$\becauseCD$是斜边$AB$中线\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\thereforeCD=BD=\dfrac\(\boldsymbol{12}\)AB$$\therefore\(BC=\)CD=BD$$\therefore\triangleBCD$是等边三角形。2.解：$\becausePC\perpOB$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\therefore\triangleOCP$是直角三角形\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\angleOCP=90^\circ$$\because\angleAOB=30^\circ$\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)$\(PC=5\)$$\thereforePC=\dfrac\(\boldsymbol{12}\)OP$$\thereforeOP=2PC=\boldsymbol{10}$##易错警示1.前提条件**直角三角形**不能丢！选择填空最爱挖坑；2.一定是**$30^\circ$角所对的直角边**\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)不是邻边；3.经常和斜边上中线一起综合出题；4.逆用：直角三角形\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)直角边$=\dfrac\(\boldsymbol{12}\)$斜边$\Rightarrow$该边对角$30^\circ$。下一节：**5.2勾股定理**\(\because\angleC=90^\circ\(\because\angleC=90^\circ\(\because\angleACB=90^\circ，\angleA=30^\circ\)\angleB=30^\circ\)\angleA=30^\circ\)继续同模板出题吗？湘教版（新教材）数学八年级上册5.1.2含30°角的直角三角形的性质授课教师：.

班

级：

8年级（

）班.

时

间：.

2026/10/10

学子如同向日葵向往暖阳，满心热爱渴求知识。在书页间汲取力量，于思索中追逐光亮，不惧学习路上的磨砺，心怀憧憬奋力向上，向着理想不断成长。第5章直角三角形123探索并掌握含30°角的直角三角形的性质（重点）掌握“在直角三角形中，如果有一条直角边等于斜边的一半，那么这条直角边所对的角等于30º.”（重点）利用“有一个角为30°的直角三角形的性质与逆命题”开展实际应用.（重点）导入新课1、直角三角形有哪些性质？结合图形，用图形语言叙述。Rt∆ABC中，∠C=90°，D是AB的中点DCBA∠A+∠B=90°CD=AD=BD=AB122、一个三角形应满足什么条件才能是直角三角形?(1)有一个角是直角的三角形是直角三角形；

(2)有两个角互余的三角形是直角三角形；（3）一边上的中线等于这条边的一半的三角形是直角三角形。03新知探究任务一：画一个锐角为30°的直角三角形用直尺分别测量斜边和直角边的长度.任务二：将直角三角形剪裁下来，将直角三角形进行折叠，使点A与点B重合，找到AB的中点D.任务三：将直角三角形进行折叠，使BC与BA所在直线重合.03新知探究思考在一个锐角为30°的直角三角板中，这个锐角所对直角边的长度与斜边的长度存在怎样的数量关系？

思考：你能通过几何语言进行证明吗？03新知探究

03新知探究直角三角形的性质3：在直角三角形中，如果一个锐角等于30°，那么它所对的直角边等于斜边的一半.

03新知探究思考可以运用轴对称知识证明结论成立吗？试一试.答案:可以.下面我们用三角尺验证:如图所示，把两个含30°角的三角尺较长的直角边拼在一起，可以构成等边三角形，该等边三角形是轴对称图形，较长的直角边所在的直线为该等边三角形的一条对称轴。这时，两条较短的直角边之和为等边三角形的一边，而斜边也